What is derivative of x^3+3x^4y=4y^2-9 ?

The answer to derivative of x^3+3x^4y=4y^2-9 is (dy)/(dx)=(-3x^2-12x^3y)/(3x^4-8y)

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derivative of x^3+3x^4y=4y^2-9

derivative of

x^{3} + 3 x^{4} y = 4 y^{2} - 9

\frac{d y}{d x} = \frac{- 3 x ^{2} - 12 x ^{3} y}{3 x ^{4} - 8 y}

Solution steps

x^{3} + 3 x^{4} y = 4 y^{2} - 9

Treat

y

y (x)

Differentiate both sides:

3 x^{2} + 3 (4 x^{3} y + \frac{d y}{d x} x^{4}) = 8 y \frac{d y}{d x}

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{- 3 x ^{2} - 12 x ^{3} y}{3 x ^{4} - 8 y}

\frac{d y}{d x} = \frac{- 3 x ^{2} - 12 x ^{3} y}{3 x ^{4} - 8 y}

100 x y + y^{2} = 5000000

\frac{sin ( x y ) + x}{y} = x^{2} - y

ln (x^{6} + y^{20}) = sin (x) + e^{y}

x y^{2} - x^{2} y - 2 = 0

x + y = x^{2} y^{2}

What is derivative of x^3+3x^4y=4y^2-9 ?
The answer to derivative of x^3+3x^4y=4y^2-9 is (dy)/(dx)=(-3x^2-12x^3y)/(3x^4-8y)